Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $k = \dfrac{x^2 - 6x}{x^2 - 3x - 18} \div \dfrac{8x + 64}{x + 3} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{x^2 - 6x}{x^2 - 3x - 18} \times \dfrac{x + 3}{8x + 64} $ First factor the quadratic. $k = \dfrac{x^2 - 6x}{(x + 3)(x - 6)} \times \dfrac{x + 3}{8x + 64} $ Then factor out any other terms. $k = \dfrac{x(x - 6)}{(x + 3)(x - 6)} \times \dfrac{x + 3}{8(x + 8)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ x(x - 6) \times (x + 3) } { (x + 3)(x - 6) \times 8(x + 8) } $ $k = \dfrac{ x(x - 6)(x + 3)}{ 8(x + 3)(x - 6)(x + 8)} $ Notice that $(x - 6)$ and $(x + 3)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ x(x - 6)\cancel{(x + 3)}}{ 8\cancel{(x + 3)}(x - 6)(x + 8)} $ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ $k = \dfrac{ x\cancel{(x - 6)}\cancel{(x + 3)}}{ 8\cancel{(x + 3)}\cancel{(x - 6)}(x + 8)} $ We are dividing by $x - 6$ , so $x - 6 \neq 0$ Therefore, $x \neq 6$ $k = \dfrac{x}{8(x + 8)} ; \space x \neq -3 ; \space x \neq 6 $